The title says it all. The answers say:

$x\le -\sqrt{3}$ and $0\le x\le \sqrt{3}$

(can someone edit this so all the $<$ have an 'or equal to' sign. Edit the roots as well please.

I'm not sure how to attempt this question. When I simplify, I get $x^2\le 3$, so $x\le \pm\sqrt{3}$.


  • 1
    $\begingroup$ For $\le$ you want \le; the plus-or-minus sign is \pm, and the square root is \sqrt{3}. $\endgroup$ – Brian M. Scott May 23 '13 at 9:13

The equality can be solved very easily.

$$x^3< 3x\iff x(x^2-3)< 0 $$

If $x<0, x^2-3>0$

As $x^2-3>0\implies$ either $x>\sqrt3$ or $x<-\sqrt3$

and as $x<0$ the required region will be $x<-\sqrt3$

Similarly, if $x>0, x^2-3<0\implies -\sqrt3<x<\sqrt3$

and as $x>0$ the required region will be $0<x<\sqrt3$



$$x^3< 3x\iff x(x-\sqrt3)(x+\sqrt3)< 0 $$

This will hold true if odd number (one or three) of factors $< 0$

Now check for the ranges $(-\infty,-\sqrt3);[-\sqrt3,0);[0,\sqrt3);[\sqrt3, \infty)$

  • $\begingroup$ I got: $x \le 0$, $x \le -\sqrt {3}$, $x \le \sqrt {3}$ What did I do wrong? $\endgroup$ – missiledragon May 23 '13 at 9:16
  • $\begingroup$ @missiledragon, how have you simplified ? Please find the edited answer $\endgroup$ – lab bhattacharjee May 23 '13 at 9:22
  • $\begingroup$ Alright, if there was a similar question, but it had an $x^2$ instead of $x^3$ would you need to test for both greater than or less than $0$? $\endgroup$ – missiledragon May 23 '13 at 9:27
  • $\begingroup$ @missiledragon, $a\cdot b<0\implies$ either $(a>0$ and $b<0)$ or $(b>0$ and $a<0)$ So, try to prove $(x-a)(x-b)<0$ where $a<b\implies a<x<b$ $\endgroup$ – lab bhattacharjee May 23 '13 at 9:28
  • $\begingroup$ There seems to be a mistake in the first explanation. If $x<-\sqrt{3}$ or $x>\sqrt{3}$, and it's known that $x<0$, then $x<-\sqrt{3}$, not $-\sqrt{3}<x<0$. $\endgroup$ – Glen O May 23 '13 at 9:35

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